1 The diagram shows a mitochondrion drawn from an electron micrograph.
X
Y
The actual length of the mitochondrion, using the line X–Y, is 3000 nm.
What is the magnification of the drawing of the mitochondrion?
A 100 B 1000 C 10 000 D 100 000

2 A specimen of plant tissue is observed twice with a microscope, firstly using red light with a
wavelength of 650 nm and then using green light with a wavelength of 510 nm.
What happens to the magnification and resolution when using green light compared to red light?
magnification resolution
A decreases decreases
B increases increases
C remains the same decreases
D remains the same increases

3 The electron micrograph shows a structure found in the cytoplasm of an animal cell.
What is this cell structure?
A centriole
B lysosome
C ribosome
D vesicle

4 Which cell structures contain nucleic acid?
1 Golgi body
2 lysosome
3 mitochondria
4 ribosomes
A 1, 2 and 3 B 1, 2 and 4 C 1 and 4 only D 3 and 4

5 Which statements about mitochondria, chloroplasts or prokaryotes are correct?
1 Mitochondria and chloroplasts have a fully permeable inner membrane and a
partially permeable outer membrane.
2 Prokaryotes and chloroplasts have 70S ribosomes that are the sites for translation
and polypeptide synthesis.
3 Prokaryotes and mitochondria have an outer membrane and an inner, folded
membrane where ATP synthesis occurs.
4 Prokaryotes and mitochondria have circular DNA where genes coding for
information are located.
A 1, 2 and 3 B 1, 2 and 4 C 1 and 3 only D 2 and 4 only

6 The very large (1000 nm) Pandora viruses found in Chile and Australia are considered to be
viruses because they cannot replicate their own genome and cannot make proteins.
They also share essential structural features with other viruses.
What are the essential structural features of viruses?
1 non-cellular
2 protein coat
3 both DNA and RNA
4 either DNA or RNA
A 1, 2 and 3 B 1, 2 and 4 C 1 and 3 only D 2 and 4 only

7 A sample of food was heated with Benedict’s solution which changed colour to green.
A second sample of the same food was boiled with dilute hydrochloric acid and neutralised using
sodium hydrogencarbonate. It was then heated with Benedict’s solution which changed colour to
red.
What did these results show?
glucose present
reducing
sugar present
non-reducing
sugar present
A
C B
D

8 Which pair of molecules only includes macromolecules that can be found in animal cells?
A amylase and amylopectin
B collagen and glycogen
C deoxyribose and starch
D sucrose and haemoglobin

9 Homogalacturonan is a polysaccharide found in plant cell walls.
The diagram shows a molecule of the monomer used to form homogalacturonan.
CO2H
HO
H
H
OH
O
H OH
OH H
A student studied the structure of this monomer and compared it with the structure of the
monomer used to form cellulose.
Which carbon atoms in the monomer in the diagram have hydroxyl groups arranged in different
positions to those found in the cellulose monomer?
A carbon one and carbon four
B carbon one only
C carbon three and carbon four
D carbon three only

10 Which statement is correct for triglycerides and phospholipids?
A A phosphate group is joined to a glycerol molecule.
B Hydrocarbon chains may be saturated or unsaturated.
C They are polar molecules.
D They contain three ester bonds.

11 Which description of collagen is correct?
A A collagen molecule consists of three polypeptide chains, each in the shape of a helix. The
three chains are wound together into a triple helix called a fibre.
B A collagen molecule consists of three polypeptide chains, each of which is an -helix. The
three chains are wound tightly together into a triple helix. Many of these triple helices bind
together as a fibre.
C A collagen molecule consists of three polypeptide chains wound tightly into a triple helix
called a fibre.
D A collagen molecule consists of three polypeptide chains in which every third amino acid is
glycine. The three polypeptides are wound tightly together into a triple helix. Many of these
helices form a fibre.

12 In a healthy human, the mean value for the number of haemoglobin molecules in one red blood
cell is 260 million.
How many -globin chains does one red blood cell contain in a healthy human?
A 1.3  108 B 2.6  108 C 5.2  108 D 1.04  109

13 Which statements about a peptide bond are correct?
1 It joins two monomers which are always identical to each other.
2 It contains four different atoms.
3 It can be broken by the addition of water at room temperature.
4 It is important in the primary structure of proteins.
A 1, 2 and 3 B 1 and 3 only C 2, 3 and 4 D 2 and 4 only

14 Which graph correctly shows the activation energy of a reaction when an enzyme is added?

15 The enzyme lactase is found in the membranes of epithelial cells lining the small intestine.
The enzyme is formed by a single polypeptide that folds to give three regions.
 an active site with the free amino group outside the cell
 a short section inside the membrane
 a short section inside the cell
What type of amino acid would be found in each of the three regions?
outside the cell inside the
membrane
inside the
cell
A hydrophilic hydrophobic hydrophilic
B hydrophilic hydrophobic hydrophobic
C hydrophobic hydrophilic hydrophobic
D hydrophobic hydrophobic hydrophilic

16 A student wrote three statements about cell signalling.
1 A signal chemical always has the same shape as a protein receptor on a target cell.
2 An increase in temperature may decrease the effect of cell to cell signalling.
3 A mutation may decrease production of active protein receptors for the cell surface
membrane.
Which statements are correct?
A 1, 2 and 3 B 1 and 3 only C 1 only D 2 and 3 only

17 At which stages of mitosis are chromosomes composed of two chromatids that are held together
by a centromere?
A anaphase and telophase
B anaphase and prophase
C metaphase and prophase
D metaphase and telophase

18 The jellyfish, Turritopsis dohrnii, is described as being immortal. If T. dohrnii is not eaten by
predators or diseased, it seems to be able to live forever. There is no way to determine the
biological age of a T. dohrnii individual.
Which feature of the cells in T. dohrnii could explain these observations?
A very long G phases in the cell cycle
B a very short S phase in the cell cycle
C an ability to restore telomeres to their original length
D fewer chromosomes than other eukaryotic organisms

19 Some chemicals, used to stop tumour growth, work by preventing the DNA double helix from
uncoiling and separating.
During which stage of the cell cycle would they act?
A anaphase
B interphase
C metaphase
D prophase

20 Four nucleotides, A, B, C and D, each consist of three phosphate groups, a nitrogenous base
and a pentose sugar. Characteristics of the base and sugar components before they are joined to
form each nucleotide are shown in the table.
Which nucleotide could pair with an adenine base during DNA replication?
ring structure of
nitrogenous base
ratio of carbon to
oxygen atoms in
pentose sugar
A double 1 : 1
B double 5 : 4
C single 1 : 1
D single 5 : 4

21 The statements describe the process of translation.
1 A peptide bond forms between adjacent amino acids.
2 Hydrogen bonds form between the anticodon and the codon.
3 mRNA binds to the ribosome.
4 tRNA enters the ribosome carrying a specific amino acid.
In which order does this process take place?
A 3  2  1  4
B 3  4  2  1
C 4  2  1  3
D 4  2  3  1

22 The sequence of amino acids in a section of a polypeptide is:
… histidine–proline–aspartic acid–leucine…
amino acid possible DNA triplet codes
aspartic acid CTA CTG
histidine GTA GTG
leucine GAT GAC
proline GGA GGG
What is a correct sequence of mRNA codons for this polypeptide section?
A …CAC CCC GAA CUG…
B …CAU CCU GAC CUA…
C …GTA CCA CTG GAT…
D …GUA GGA CUG GAU…

23 What contributes to the upward movement of water in a xylem vessel of a plant?
1 cohesion of water molecules by hydrogen bonding
2 adhesion of water molecules to the cellulose walls of xylem vessels by hydrogen
bonding
3 removal of water from xylem vessels in a leaf reduces the hydrostatic pressure in
the xylem
A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only

24 The diameter of a tree trunk usually decreases slightly during the day.
Which changes in environmental factors during the day could cause the diameter to decrease
even more?

25 The diagram shows a transverse section of a stem.
Which area is the phloem?

26 The diagram shows a xerophytic leaf in different conditions, P and Q.

Which statements describe the difference between the cells in layer Y in conditions P and Q?
1 more negative water potential in P than Q
2 more cells plasmolysed in P
3 cells less turgid in Q
4 water potential becomes zero in Q
A 1, 2 and 3 B 1 and 2 only C 2 and 4 D 3 and 4

27 Different substances, such as sucrose and amino acids, can move in different directions in the
phloem sieve tube elements.
Which statement explains this?
A Active transport occurs in some phloem sieve tube elements and mass flow occurs in other
sieve tube elements.
B Both active transport and mass flow occur in each individual phloem sieve tube element.
C Mass flow occurs in both directions at the same time in each individual phloem sieve tube
element.
D Mass flow occurs in different directions in different phloem sieve tube elements at the same time.

28 The statements list some of the events in a cardiac cycle. The statements are not in the correct
order.
Which statement describes the fourth of these events to occur in the cardiac cycle?
1 The impulse travels through Purkyne tissue.
2 A wave of excitation sweeps across the atria.
3 The atrioventricular node delays the impulse for a fraction of a second.
4 The sinoatrial node contracts.
5 The wave of excitation sweeps upwards from the base of the ventricles.
6 The ventricles contract.
7 The atria contract.
A 1 B 3 C 4 D 7

29 Which row correctly identifies components of both lymph and tissue fluid?
antibodies red
blood cells
sodium
ions
white
blood cells
A     key
B     = component present
C      = component not present
D    

30 Which row is correct for the mean blood pressure in different parts of the human circulatory
system?
mean blood pressure
right
atrium
artery in
arm
vein in
arm
capillary
in arm key
A + ++++ ++ +++ ++++ highest mean blood pressure
B ++ ++++ +++ + +++
C +++ ++ ++++ + ++
D ++++ +++ + ++ + lowest mean blood pressure

31 Which mechanism accounts for the way most of the carbon dioxide is transported in blood?
A Carbon dioxide dissolves in plasma and is carried in solution.
B Carbon dioxide is converted to carbaminohaemoglobin inside red blood cells.
C Carbon dioxide is converted to carboxyhaemoglobin inside red blood cells.
D Carbon dioxide is converted to hydrogencarbonate ions inside red blood cells.

32 The graph shows the dissociation curves for haemoglobin at two different partial pressures of
carbon dioxide.
At which position on the graph, A, B, C, or D, is the concentration of haemoglobinic acid lowest in
red blood cells?

33 The table shows the partial pressure of carbon dioxide and oxygen in two blood vessels.
partial pressure of
gas in pulmonary
artery / kPa
partial pressure of
gas in pulmonary
vein / kPa
carbon dioxide 6 5
oxygen 5 15
What explains the difference in the partial pressures of oxygen in the pulmonary artery and
pulmonary vein?
A Oxygen diffuses from the alveoli into the blood in the capillaries.
B Carbon dioxide diffuses into the alveoli from the blood in the capillaries.
C Oxygen diffuses from the body cells into the blood in the capillaries.
D Carbon dioxide diffuses into the body cells from the blood in the capillaries.

34 Which graph shows the effect of carbon monoxide on the percentage saturation of haemoglobin
with oxygen?

35 Which row shows the correct methods of transmission of the named pathogens?
method of transmission
airborne droplets insect vector water
A Morbillivirus Mycobacterium Vibrio
B Mycobacterium Plasmodium Vibrio
C Plasmodium Vibrio Morbillivirus
D Vibrio Plasmodium Mycobacterium

36 Which disease is caused by a eukaryote?
A cholera
B malaria
C measles
D smallpox

37 What is the initial mechanism by which bacteria become resistant to antibiotics?
A genetic mutation
B overuse of antibiotics
C natural selection
D patients not finishing a course of antibiotics

38 What are functions of T-lymphocytes?
1 production of cytokines
2 production of toxins
3 recognition of an antigen bound to an antigen-presenting cell
A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only

39 What is correct about the role of memory cells in long-term immunity?
1 They divide to form plasma cells and memory cells when the pathogen enters the
body a second time.
2 They produce a fast response so that the person infected with the pathogen does
not become ill again.
3 They produce more antibodies than were produced during the primary immune
response.
4 They remain in the blood and lymphatic system after the pathogen has been
destroyed.
A 1, 2 and 3 B 1, 2 and 4 C 1 and 3 only D 2 and 4 only

40 Which events result in a person developing actively acquired immunity?
1 becoming infected by TB bacteria
2 drinking breast milk
3 receiving an injection of antigens
4 receiving an injection of antibodies
A 1, 2 and 3 B 1, 3 and 4 C 1 and 3 only D 2 and 4

Ans:

Question Answer Marks
1 C 
2 D 
3 A 
4 D 
5 D 
6 B 
7 C 
8 B 
9 A 

10 B 
11 D 
12 C 
13 D 
14 D 
15 A 
16 D
17 C 
18 C 
19 B 
20 D 
21 B
22 B 
23 A 
24 D 
25 B 
26 B 
27 D 
28 B 

29 B 
30 A 
31 D 
32 C 
33 A 
34 A 
35 B 
36 B 
37 A 
38 A 
39 B 
40 C 

1 Fig. 1.1 is a transmission electron micrograph of cells from duckweed, Spirodela oligorrhiza.

(a) Calculate the actual width of the cell labelled X.
Write down the formula you will use to make your calculation.
Show your working and give your answer in micrometres to one decimal place.

Ans: actual width = image width ÷ magnification ;
A A = I ÷ M M = I ÷ A I = A
× M or magnification triangle
working = width divided by 4275 ;
e.g.
16 000 ÷ 4275 17 000 ÷ 4275 18 000 ÷ 4275 19 000 ÷ 4275 ;
3.7 (
μm) 4.0 (
μm) 4.2 (
μm) 4.4 (μm) ;
R answer if given to more than 1 dp or whole number

(b) (i) Table 1.1 lists some biological molecules found in plant cells.
Complete Table 1.1 by choosing one letter from Fig. 1.1 that indicates a cell structure
where each biological molecule is found.
Table 1.1
biological molecule letter from Fig. 1.1
DNA
cellulose
phospholipid
histone proteins
(ii) State the name of a cell structure, visible in Fig. 1.1, where ATP is synthesised.

Ans: 

DNA –
A / B / C ;
cellulose –
E ;
phospholipid –
A / C ;
histone proteins
–
A / B ;

(iii) Name a cell structure that produces mRNA.

Ans: chloroplast / mitochondrion ;

(c) Describe the evidence from Fig. 1.1 that shows that the image is a transmission electron
micrograph.

any two from
1 (section at) high resolution ;
A suggestion of a correct value of resolution for a TEM
2 any named structure visible in Fig. 1.1 that can, only be seen in a TEM / not be seen in a photomicrograph ;
e.g. internal structure of chloroplasts / thylakoid(s) / grana
e.g. internal structure of mitochondria / cristae
3 high magnification / higher magnification (magnification > 1000 / higher than with light microscope) ;
in context of higher than light microscope
4 (very) thin ;
5 2D / no surface contours / no surface features / AW ; A not 3D

2 Fig. 2.1 shows three molecules of water

(a) Describe the hydrogen bonding that occurs between the water molecules shown in Fig. 2.1.

Ans: any three from
1 (diagram shows) hydrogen bond is a weak bond ;
2 each oxygen (atom) forms two hydrogen bonds / each hydrogen (atom) forms one hydrogen bond ;
3 (attraction) between oxygen (atom) of one water molecule and hydrogen (atom) of another (forms a hydrogen bond) ;
R cohesion / adhesion – if used for attraction
4 water is dipolar ;
5 detail ;
e.g. electrons not shared equally between oxygen and hydrogen
A oxygen is more electronegative (than hydrogen)
A ref. to (two) lone pair(s) (of electrons) on oxygen
A uneven distribution of, electrons / charge
e.g. oxygen has, small / slight, negative charge / δ–, and, hydrogen has, small / slight, positive charge / δ+
only needs to state ‘small’ once

(b) The human enzyme, salivary amylase, is composed of one polypeptide. Fig. 2.2 represents
the structure of a molecule of salivary amylase.

Ans: idea that (H-bonds) maintain / AW, (shape / structure, form of)
α-helices / β-pleated sheets ;
A allows formation of, α-helices / β-pleated sheets
R if bonds are between R groups

(i) Explain the role of hydrogen bonding in maintaining the secondary structure of proteins,
such as salivary amylase.

Ans: idea that (H-bonds) maintain / AW, (shape / structure, form of)
α-helices / β-pleated sheets ;
A allows formation of, α-helices / β-pleated sheets
R if bonds are between R groups

(ii) Explain the role of hydrogen bonding in maintaining the tertiary structure of proteins such
as salivary amylase.

Ans: any two from
1 idea that hydrogen bonds help to, stabilise / AW, further folding of, amylase / polypeptide / protein ;
2 between, R groups with amine and carboxyl groups ;
A between R groups with -NH and, -CO / -OH
3 idea that may be between amino acids far apart in primary structure ;
4 either helps to maintain / form / AW, globular shape / 3D shape / structure (of amylase / polypeptide / protein)
or
maintains / forms / AW, (specific) shape / structure, of, active site / binding site ; 

(c) Outline the importance of water as a solvent in plants.

Ans: any three from
1 dissolves / AW, ions / minerals / salts, and (named) polar molecules ;
A ‘assimilates’ as polar I substances / nutrients
2 transports, solute(s) / named solutes / dissolved substance, in, xylem / phloem / xylem and phloem ;
3 storage of, solutes / named solutes, in vacuoles ;
4 metabolic / chemical / cellular, reactions occur in water ;
5 dissolves, carbon dioxide / oxygen, with ref to, respiration / photosynthesis;

3 Visking tubing can be used to investigate the properties of cell membranes.
A student carried out an experiment to investigate osmosis using Visking tubing. An outline of the
investigation is shown in Fig. 3.1.

Six pieces of Visking tubing were filled with 10cm3 of different concentrations of sucrose
solution: 0.0, 0.4, 0.8, 1.2, 1.6 and 2.0moldm–3.
• The height of the meniscus of each solution in the Visking tubing was measured.
• The pieces of Visking tubing were put into test-tubes containing 15cm3 of 0.9moldm–3
sucrose solution.
• After 20 minutes, the pieces of Visking tubing were removed from the test-tubes and the
height of the meniscus in each was measured.
The results are shown in Table 3.1.
Table 3.1
concentration of sucrose
solution inside Visking tubing
/moldm–3
difference in height of
meniscus after
20 minutes/mm
0.0 –12
0.4 –4
0.8 –2
1.2 +1
1.6 +6
2.0 +11

(a) The Visking tubing used by the student was not permeable to sucrose.
Explain the results shown in Table 3.1.

Ans: any three from
1 differences in height show that
concentrations of sucrose, 0, 0.4 and 0.8 (mol dm–3) or
⩽ / less than, 0.8 / 0.9, water moves out of Visking tubing ;
A one of 0, 0.4 or 0.8 R sucrose moving
2 concentrations of sucrose 1.2, 1.6, 2.0 (mol dm-3) or
⩾ / more than, 0.9 /1.2, water moves into Visking tubing ;
A one of 1.2, 1.6 or 2.0 R sucrose moving
3 ref. to net water movement ;
if water enters Visking tubing – A ora for water leaving
4 external solution has higher water potential (than contents of Visking tubing) ;
A high water potential to low water potential
5 water moves, down, water potential / Ψ, gradient ; 

(b) When red blood cells are placed in water they are destroyed by bursting.
The student also investigated how red blood cells are affected by immersion in solutions of
sodium chloride of different concentration. Blood samples of the same volume were added to
solutions of sodium chloride in separate test-tubes.
After 10 minutes, the student took 0.1cm3 of the blood samples from the test-tubes and
estimated the percentage of red blood cells that had burst in each blood sample.
Fig. 3.2 shows the student’s results.

Describe and explain the effects on red blood cells of immersion in different concentrations of
sodium chloride as shown in Fig. 3.2.

Ans: mark whole question to a max of four marks
descriptions
1 at concentrations, less than / ⩽, 0.04 mol dm–3 all cells, burst / AW ;
2 at concentrations between 0.04 and 0.14 mol dm–3 decreasing percentage of cells burst / AW ; A use of percentages
3 at concentrations, greater than / ⩾, 0.14 mol dm–3 no cells, burst / AW ;
explanations to max 3
4 in low concentrations / ⩽ 0.04, of sodium chloride water moves into cells down water potential gradient / from high
Ψ to
low
Ψ ;
5 cells increase in, volume / size / internal pressure ;
6 either
cell membranes are not strong enough to withstand increase in volume / pressure
or
red blood cells burst because they have no cell wall ;
7 between 0.04 and 0.14 (mol dm–3) water potential gradient into cells, decreases / becomes less steep / AW ;
8 above 0.14 (mol dm–3) / in high concentrations, water potential inside cells is the same or higher than the sodium
chloride solution ;
9 at high concentration / ⩾ 0.14, water leaves cells / cells shrink / cells shrivel / cells show crenation ; 

4 (a) The induced-fit hypothesis and the lock-and-key hypothesis are used to describe the mode of
action of enzymes.
Explain the induced-fit hypothesis.

Ans: idea of shape needs to be in the answer for mp2 and mp4 I ‘substrate changes shape’
any three from
1 active site is not (fully) complementary to substrate ;
2 active site, changes shape / moulds around, to fit the substrate ;
A conformational change for shape change
3 enzyme-substrate complex / ESC, forms ;
4 active site returns to original shape on release of product ;
5 AVP ;
e.g. change of shape (to give complementary fit) lowers activation energy / puts strain on bonds / AW
ref. to binding site / catalytic site (of active site)

(b) Radish plants contain the enzyme peroxidase that catalyses the breakdown of hydrogen
peroxide.
Students investigated the effect of increasing the concentration of hydrogen peroxide on the
activity of peroxidase extracted from radish.
The results of their investigation are shown in Fig. 4.1.

(i) Explain the effect of increasing the concentration of hydrogen peroxide on the initial rate
of reaction as shown in Fig. 4.1

Ans: A peroxidase for enzyme
any three from
rate of reaction increases as substrate concentration increases to 0.33 – 0.35 mmol dm-3
1 (up to 0.33 to 0.35 mmol dm–3) some active sites are not occupied ;
2 more collisions between, enzyme / active site, and substrate molecules ;
A more collisions lead to increase in formation of enzyme-substrate complexes
rate of reaction remains constant above 0.33–0.35 mmol dm–3
3 all active sites, occupied / saturated ;
4 correct reference to limiting factor for slope ; A limits / limiting
e.g. at low substrate concentrations, substrate concentration limiting / enzyme concentration not limiting
5 correct reference to limiting factor for plateau ; A limits / limiting
e.g. at high substrate concentrations, enzyme concentration limiting / substrate concentration not limiting

(ii) The students determined the Km for radish peroxidase as 0.10mmoldm–3.
With reference to Fig. 4.1, describe how they determined the Km.

Ans: idea that determined Vmax / maximum rate / ref. to 5.6 (
μmol min–1) ;
A if Vmax shown on graph
K
m is the substrate concentration at half the Vmax ;

(c) A further investigation found that the Km for carbonic anhydrase is 12mmoldm–3.
Describe the role of carbonic anhydrase in the transport of carbon dioxide.

Ans: any three from
in respiring tissue
1 (carbonic anhydrase) catalyses the reaction between water and carbon dioxide in, red (blood) cells / erythrocytes ;
2 to form carbonic acid, which dissociates to form HCO3
– and H+ ;
A from equation
A use of word ion(s)
3 HCO3
–, moves / passes / diffuses (through membrane proteins from red blood cells) into plasma ;
4 (activity of enzyme) maintains (steep) concentration gradient for diffusion of carbon dioxide into red blood cells ;
5 idea that so helps to remove large quantities of carbon dioxide ;
in lungs
6 (carbonic anhydrase) converts HCO3
– back into carbon dioxide so can be, excreted / removed / exhaled ;
if described earlier
A catalyses reverse reaction so carbon dioxide can be, excreted / AW 

5 Much success was made in reducing the number of cases of malaria between 2000 and 2015.
(a) Explain how malaria is transmitted.

Ans: 1 vector is female Anopheles (mosquito) ;
2 mosquito / (female) Anopheles, takes blood from infected person ; I ‘bites’ alone
3 (vector / mosquito / Anopheles) inserts / AW, saliva / anticoagulant, with, pathogen / Plasmodium / parasite, into (blood
of) uninfected person ;
R if incorrect type of pathogen

(b) Diagnostic test strips for malaria contain monoclonal antibodies. The test strips detect
antigens produced by the pathogens that cause malaria.
Fig. 5.1 shows stages in the production of monoclonal antibodies. The information in three of
these stages is incomplete.
Complete Fig. 5.1.

Ans: B-lymphocytes / B-cells ;
fuse / combine / join / merge / AW ;
R hybridise / bind / bond
hybridoma ; 

(c) Fig. 5.2 shows two diagnostic test strips for malaria.

A sample of blood from a person suspected of having malaria is put into the well
labelled S.
• A buffer solution is put into the well labelled A.
• The buffer solution moves the blood towards the results window.
• A line at position C indicates that the test is working correctly.
• A line at position T indicates a positive result for malaria.
State three advantages of using test strips for malaria, such as those shown in Fig. 5.2.

Ans: any three from
1 results are accurate (rely on monoclonal antibodies) ;
2 results are obtained quickly / diagnosis is quick ;
3 low cost / cheap ;
4 does not need highly trained medical professional / easy to read / easy to use / AW ;
5 does not need specialised equipment ;
6 easily, transported / distributed (in bulk) ;
7 can ensure appropriate treatment is given immediately malaria is confirmed ;
8 reduces unnecessary use of antimalarial drugs (just in case) ;
9 (a small area for S so) only a small sample of blood needed ;
10 AVP ;
e.g. can identity different species of Plasmodium
I ref. to line C and working correctly

(d) The highest number of cases of malaria occur in sub-Saharan Africa and South-East Asia.
Discuss the factors that determine the global distribution of malaria.

Ans: ignore ref. to sickle cell anaemia
any five from
habitat of Anopheles mosquito
1 tropical / warm and humid / AW, climate / regions ;
A in context of global warming (increases life span of Anopheles)
2 in areas where, Anopheles (mosquito) / vector, occurs ;
3 Anopheles mosquitoes only live in humid conditions ;
4 warm temperatures for, development / growth, of, parasite / Plasmodium (in vector / mosquito) ;
5 warm temperature for, development / growth, of mosquito larvae ;
6 mosquitoes require bodies of, still / AW, water for breeding ;
A ponds, puddles, lakes, swamps
7 mosquitoes require places where there is sufficient rainfall ;
8 low altitude ;
resistance
9 insecticide / repellent, resistance of mosquitoes ;
10 drug resistance of parasite ;
prevention
11 in countries / areas, where, prevention / control measures, are not implemented by, governments / health authorities ;
12 further detail e.g. any example of a, prevention / control measure that is not used or not implemented fully ;
immunity
13 immunity to malaria in human population (limits distribution) ;
people
14 migration of infected people from areas (with high rates of malaria) ;
15 high rates of HIV infection ;

6 Lysosomes are cell structures that contain enzymes known as acid hydrolases.
Fig. 6.1 shows some processes that occur in animal cells

(a) Name the cell structures labelled A and E

A rough endoplasmic reticulum ;
E Golgi, body / apparatus / complex ;

(b) State the function of the structures labelled F.

Ans: transport from, (R)ER / ribosomes, to Golgi ; A from
A to E
A transport from Golgi to, cell surface membrane / phagosome
A separate, acid hydrolases / enzymes, from, rest of cell / AW 

(c) Name the process by which bacteria are taken into the cell at C.

Ans: phagocytosis / endocytosis ; 

(d) With reference to the processes occurring at B and at D in Fig. 6.1, outline the role of acid
hydrolases in lysosomes.

Ans: any three from
1 break down / digest / destroy, bacteria / pathogen(s) ;
2 break down / digest / destroy, (worn out / defective / AW), organelles / named organelle (in animal cell) ; A autophagy
3 catalyse / speed up, hydrolysis ;
4 any two named substrates ;
e.g. (any named) polysaccharides / proteins / (phospho)lipids / (named) nucleic acids
5 idea that recycle / reuse, biological molecules within cell ;
6 (macrophage / phagocyte) cut up to present antigen ; 

(e) Carrier proteins in the membranes of lysosomes maintain a lower pH than the surrounding
cytoplasm by moving hydrogen ions.
Suggest how the carrier proteins maintain the lower pH within the lysosomes.

Ans: moves / pump(s), hydrogen ions / protons, into the lysosome against concentration gradient ;
active transport / uses ATP / energy from respiration / ref to conformational change of carrier ;
R if a ref to facilitated diffusion 

1 (a) Photosynthesis is a complex process involving the transfer of light energy into chemical
energy.
(i) Describe the role of photosynthetic pigments.

Ans: any two from:
1 absorb / harvest / capture / trap, light / photons ;
2 (to) excite electrons ;
3 (for, cyclic / non-cyclic) photophosphorylation / electron transport chain ;
4 accessory pigment passes energy to, primary pigment / reaction centre ;
5 (pigments) form a, light harvesting cluster / photosystem / PSI / PSII ; 

(ii) Name the precise location in a chloroplast of photosynthetic pigments.

Ans: thylakoid membrane(s) / granum / grana / lamella(e) ; 

(iii) Name a practical technique to separate photosynthetic pigments.

chromatography ;

(b) The rate of photosynthesis is affected by many environmental factors.
A student carried out an experiment to investigate the relationship between the concentration
of carbon dioxide available to an aquatic plant, Cabomba caroliniana, and its rate of
photosynthesis.
Fig. 1.1 shows the experimental set-up for this investigation.

Sodium hydrogencarbonate solution was used as a source of carbon dioxide.
• The concentration of carbon dioxide was varied using six different concentrations of
sodium hydrogencarbonate solution.
• All C. caroliniana plants were kept in the dark before the light was switched on at
the start of the experiment.
• Five replicates were carried out at each concentration.
• The rate of photosynthesis was obtained by calculating the percentage change in
dissolved oxygen concentration in the solution over five minutes.
Fig. 1.2 shows the results of the investigation.

(i) With reference to Fig. 1.2, explain the pattern of results obtained between 0.25moldm–3
and 1.25moldm–3 of sodium hydrogencarbonate solution.

Ans: 1 (from 0.25) to 1.00 (mol dm–3), concentration / CO2, is limiting factor
or
at / >, 1.00 (mol dm–3), concentration / CO2, no longer limiting factor ;
2 at / >, 1.00 (mol dm–3), light intensity / temperature, is limiting factor ;
3 CO2, needed / used, for Calvin cycle / for light independent stage
or
CO2, reacts with / fixed by, RuBP ; 

(ii) The percentage change in dissolved oxygen for C. caroliniana at 0.00moldm–3 of sodium
hydrogencarbonate solution is negative.
Suggest reasons for this negative value.

Ans: little / no, photosynthesis ;
(aerobic) respiration uses oxygen ; 

(iii) To minimise temperature changes, the student decided to use an LED lamp as a light
source. LED lamps release very little heat energy.
Explain the importance of minimising temperature changes in this experiment

Ans: any three from:
1 to control a variable / to ensure a fair test / so CO2 is the only variable ;
2 low, temperature / kinetic energy, decreases, rate / reaction / photosynthesis / O2 production / O2 increase ;
3 low, temperature / kinetic energy, decreases, (named) enzyme-substrate, collisions / complexes ;
4 low(er) temperature, is a limiting factor / may limit rate ;
5 high, temperature / kinetic energy, decreases, rate / reaction / photosynthesis / O2 production / O2 increase ;
6 high, temperature / kinetic energy, denatures (named) enzyme(s) ;
7 high / increased, temperature, causes / increases, photorespiration ; 

2 (a) All organisms respire to release energy from energy-rich molecules such as glucose.
(i) Anabolic reactions use energy.
State two examples of anabolic reactions occurring in living organisms.

Ans: any two from:
idea of synthesis from smaller units
1 DNA replication / transcription ;
2 to, make / build up, (named) protein ;
3 to, make / build up, (named), lipid / triglyceride / phospholipid ;
4 to, make / build up, (named) carbohydrate / polysaccharide ;
5 phosphorylation ; 

(ii) Describe and explain the use of ATP in glycolysis.

Ans: any three from:
1 phosphorylates glucose
or
glucose
 glucose / fructose / hexose, phosphate ;
2 second phosphorylation
or
glucose / fructose / hexose, phosphate
 fructose / hexose, bi(s)phosphate ;
3 so glucose cannot leave the cell ;
4 activates / increases energy of / decreases stability of, glucose ; 

(b) Some factory workers in the early 20th century were exposed to chemical X and experienced
serious side-effects.
Some of the effects of exposure to chemical X are:
• decreased production of ATP
• increased lipid metabolism, with weight loss
• increased production of pyruvate and lactate
• excess heat energy release, causing an increase in body temperature, which can
be fatal.
Chemical X increases the permeability of the inner mitochondrial membrane to protons
(hydrogen ions), causing some protons to leak out into the matrix.

(i) Explain why people exposed to chemical X show:
• decreased production of ATP
• increased lipid metabolism.

Ans:  any three from:
decreased production of ATP
1 smaller / less steep, proton gradient ;
2 few(er) protons go through, ATP synth(et)ase / stalked particle ;
increased lipid metabolism
3 use / burn / respire, body fat / lipid stores ;
4 as less ATP made from, glucose / food ; 

(ii) Suggest and explain why chemical X causes increased production of pyruvate and
lactate

Ans: any two from:
1 more / faster / increased, (rate of) glycolysis ;
2 oxygen, decreased / used faster / shortage ;
3 (so) more / increased, anaerobic respiration ;

(iii) Suggest why excess heat energy is released.

Ans: any one from:
energy from, reduced NAD / ETC, released as heat ;
energy not used to join ADP and Pi so released as heat ; 

3 In 2014, scientists published a phylogenetic (evolutionary) tree diagram showing how and when
different groups of insects developed from ancestral groups.
To construct the diagram, the scientists compared nucleotide and amino acid sequence data for
1478 genes from each of 103 different insect species. The 103 species represented all 32 orders
of the class Insecta.
The data were used to arrange the species in related groups according to their degree of molecular
similarity and to track how sequences changed over time.
(a) (i) Outline three ways in which the members of a species are similar.

Ans: any three from:
(individuals
/ organisms
/ members of a species are similar
/ same)
1 morphologically ;
2 physiologically ;
3 biochemically ;
4 genetically ;
5 ecologically / occupy same niche ;
6 behaviourally ;
7 can (inter)breed to form fertile offspring ;
8 reproductively isolated (from other species) ;
9 share, an evolutionary lineage / recent common ancestor ;
10 share a, binomial / Linnaean, name ; 

(ii) Organisms are classified in a hierarchical system of taxonomic units.
Name the domain and kingdom to which insects belong.

Ans: domain: Eukarya ;
kingdom:
Animalia ;

(iii) Explain why a phylogenetic tree diagram is useful for classifying insect species.

Ans: any two from:
1 classification / taxonomy, reflects / shows / depends on, evolutionary / phylogenetic, relationships ;
2 organisms / species, that, are closely related / share recent common ancestor, should be classified together ;
3 help identify, larger taxonomic groups / orders / families / genera ;
4
AVP ;

(b) For this study, new software for processing large quantities of data was developed.
(i) Calculate the total number of sequences that were compared, considering both the
nucleotide and amino acid sequences for each gene.

Ans: (1478
× 103
× 2 =) 304 468 ;

(ii) Explain why this study needs to make use of bioinformatic techniques.

Ans: any three from:
1 to, store / analyse / process ;
2 large quantity of data ;
3 to, compare / match, (many) sequences ;
4 to, calculate / measure / quantify, (percentage), similarity / difference ;
5 ref. to fast / accurate / efficient ;

4 (a) Genetic engineering involves the manipulation of naturally occurring enzymes and processes.
List the enzymes used in genetic engineering (genetic modification) and outline their roles in
natural processes.

Ans: any five from:
1 (named) restriction, enzyme / endonuclease ;
2 cut / fragment, DNA ;
3 produced by / present in, bacteria / archaea / prokaryotes
or
to defend against, viruses / (bacterio)phages ;
4 (DNA) ligase ;
5 joins DNA / forms phosphodiester bonds / seals sugar-phosphate backbone ;
6 during DNA, repair / replication
or
joins Okazaki fragments ;
7 reverse transcriptase ;
8 forms / makes, (c)DNA from mRNA (template) ;
9 produced by / present in, retroviruses ;
10 DNA / Taq, polymerase ;
11 replicates / copies, DNA (at high temperatures) ;
12 produced by / present in, Thermus aquaticus ; 

(b) Genetically modified organisms (GMOs) include crop plants. Genetically modified (GM) crop
plants have been grown in North America in increasing quantities since 1996. They are now
grown in many areas of the world and are eaten by millions of people and farm animals.
Table 4.1 compares the area of land used to grow GM crop plants on six continents in 2013.
Table 4.1
continent
area of land used to
grow GM crop plants
/ million hectares
edible GM crop plants grown
Africa 3.5 maize, soybean
Asia 19.1 maize, papaya, tomato
Australasia 0.6 canola
Europe 0.2 maize
North America 81.1 canola, maize, papaya,
soybean, squash, sugar beet
South America 70.9 canola, maize, soybean

(i) Suggest how North America and South America benefit from growing large areas of GM
crop plants rather than non-GM crop plants.

Ans: any three from:
1 increased / higher / more, yield / production / harvest / food ;
2 increased / higher / better, quality of, produce / crops / food ;
3 decreased / lower / less, insecticide / pesticide, needed / used ;
4 decreased / lower / less, fuel use / CO2 emissions ;
5 potential for, set-aside / conservation areas (as less land needed for crops) ;
6 (described) comparative economic benefit ; 

(ii) Only a small area of land is used to grow GM crop plants in Europe. This is mainly
because most of Europe uses a ‘precautionary principle’. The precautionary principle
prevents GM crop plants being grown if there is a possible risk of harm to human health
or to the environment, even if there is no proof of harm.
Many European citizens have concerns about the safety of eating GM food, but Europe
imports large quantities of GM maize and GM soybean to feed farm animals.
Use this information and Table 4.1 to deduce two arguments in favour of growing more
GM crop plants in Europe.

Ans: any two from:
1 people in other (named) continents eat GM, food / crops ;
2 GM, food / crops, already in (European) food chain ;
3 ref. to economic benefit of reducing GM (animal food) imports ; 

(iii) Explain why the data in Table 4.1 are not enough to calculate the extent to which different
continents have replaced traditional crop plants with GM versions.

Ans: 1 do not know / need to know, total crop area / non-GM crop area ;
2 so can work out, percentage / fraction, of, GM / non-GM ;

5 (a) Nerves consist of bundles of different types of neurones.
Fig. 5.1 shows the mean transmission speed of impulses for three types of sensory neurone,
C, Aδ and Aβ.

Impulses from pain receptors in the skin are sent along sensory neurones C and Aδ.
• Sensory neurones C and Aδ synapse in the spinal cord with relay neurones known
as projection neurones.
• Projection neurones send impulses to the part of the brain that perceives pain.
• Impulses from touch receptors in the skin pass along sensory neurones Aβ, which
can also synapse with the projection neurones in the spinal cord.
• Sensory neurones Aδ and Aβ are myelinated.
• Sensory neurones C are unmyelinated.

(i) Describe the results shown in Fig. 5.1 and suggest explanations for these results.

Ans: any three from:
1 wider / thicker / larger diameter
 increases / higher, speed ;
2 large(r) membrane surface area so more ion movement ;
3 myelin
 increases / faster / higher, speed ;
4 saltatory conduction (in myelinated) ;
5 pairwise data comparison ;

(ii) Gently rubbing a damaged area of skin can reduce the perception of pain.
Suggest an explanation for this.

Ans: any one from:
1 impulses, for touch / in
Aβ neurones, travel faster ;
2 (Aβ neurones) inhibit, spinal cord / relay / projection, neurones ;
3
AVP ;

(b) Back pain in humans can be reduced by using a transcutaneous electrical nerve stimulation
(TENS) machine.
A TENS machine uses electrical impulses to stimulate nerve endings near the site of the
pain.
It is thought that TENS triggers the release of natural painkillers called endorphins at
synapses.
Fig. 5.2 shows synapses in a pain pathway.

(i) Suggest and explain how endorphins may act to reduce pain.

Ans: any four from:
1 endorphins bind to (endorphin) receptors ;
2 stop, calcium ions / Ca2+, entering presynaptic knob ;
3 no / fewer, vesicles, move towards / fuse with, (presynaptic) membrane ;
4 no / less, acetylcholine / ACh, released ;
5 no / less, binding of ACh to postsynaptic receptors ;
6 no / less, depolarisation of postsynaptic, neurone / membrane ;
7 no / fewer, action potentials / impulses, to, pain centre / brain ; 

(ii) Suggest one disadvantage of using pharmaceutical drugs for reducing pain compared
to using TENS.

Ans: any one from:
1 (drug) dependency / addiction / tolerance ;
2 (drug) side-effects ;
3 expensive ;
4 work more slowly ; 

6 (a) Thermoregulation is the control of the core temperature of the body.
Thermoreceptors send information about a change in core temperature to the hypothalamus.
The hypothalamus controls the core temperature by sending impulses to activate several
physiological responses.
Table 6.1 lists physiological responses to a decrease in core temperature.
Complete Table 6.1 to explain how each response contributes to maintaining a core
temperature within narrow limits.

response explanation
blood vessels near the surface of the skin constrict
(narrow) less heat lost as, less / slower, blood (flow) at, skin / surface ;
shivering occurs muscle contraction, creates / generates / produces, heat ;
hairs rise trapped air, insulates / reduces thermal gradient ;
reduction in the production of sweat to stop / decrease, heat loss by evaporation ;

(b) Animals can also use behaviour to adapt to temperature changes.
Suggest why some animals curl up their bodies in cold weather.

Ans: decrease / less, surface area to, conserve / retain heat
or
decrease / less, surface area so less heat lost ; 

(c) The Sitatunga deer, Tragelaphus spekii, and the Nile monitor lizard, Varanus niloticus, live in
Africa.
Fig. 6.1 shows a Sitatunga and Fig. 6.2 shows a Nile monitor.

An investigation was carried out to measure the core temperature of the Sitatunga and the
Nile monitor as the environmental temperature was increased.
The results are shown in Fig. 6.3.

(i) Describe the results shown in Fig. 6.3.

Ans: any three from:
1 Sitatunga’s (core / body) temperature (fairly) constant ;
2 Nile monitor’s / lizard’s, (core / body) temperature rises as environmental temperature rises ;
3 figures for two animals at one temperature with units
or
one animal at two temperatures with units ;
4
AVP ;

(ii) When the environmental temperature is low, Nile monitors are relatively inactive.
Suggest one disadvantage to the Nile monitor of being inactive.

Ans: problem with, catching prey / feeding / avoiding predation ;

(d) Thermoregulation relies on negative feedback.
Explain what is meant by negative feedback.

Ans: 1 ref. to set point / norm(al) / ideal value ;
2 rise / fall / change / fluctuation ;
3 reversed / corrected / counter-acted ; 

7 People vary in which molecules are present on the surface of their red blood cells. For example, in
the ABO blood group system, people may be categorised into one of four blood groups: A, B, AB
and O.
A single gene, I, with three alleles, determines these blood groups. Gene I is an example of a
gene with multiple alleles.
• Alleles IA and IB are codominant.
• Allele Io is recessive to alleles IA and IB.
(a) Explain what is meant by codominant and recessive

Ans: codominant
1 both (alleles), affect / seen in / expressed in / contribute to (phenotype) ;
2 in heterozygote ;
recessive
3 affects / seen in / expressed / contributes to, (phenotype) (only) when dominant (allele) absent / in homozygote ;

(b) Table 7.1 shows a cross between a person with blood group A and a person with blood
group B.
Complete Table 7.1.

Ans: 

parental phenotype
A B
parental genotype IA Io IB Io
gametes I
A Io IB Io ;
offspring genotype I
A IB IA Io IB IO Io Io ;
offspring phenotype
AB
A B O ;

(c) Haemophilia is a blood condition where bleeding continues for a long period of time after a
blood vessel has been damaged.
A person with haemophilia has a mutation in a gene on the X chromosome, which results in
the lack of a blood clotting factor.
Explain why a man who has haemophilia cannot pass the condition to his male offspring.

Ans: any three from:
1 males / men / boys, are, XY / have (only) one X chromosome ;
2 man / father, passes Y to, male offspring / son ;
3 man / father, does not pass X to,
♂ offspring / son ;
4 man / father, passes, X / mutation, to, female offspring / daughter ;
5 woman / mother, passes X / mutation, to, male offspring / son ; 

8 (a) Humans use selective breeding to improve the features of crop plants and domestic
animals.
(i) State another term for selective breeding.

Ans: artificial selection ;

(ii) Describe how selective breeding can be used to improve the milk yield of dairy cattle.

Ans: any five from:
1 (humans) choose / cross / breed from, females / cows / cattle, with, good / high, milk yield ;
2 (humans) choose / cross / breed from, males / bulls, with, mother / sisters / daughters, with, good / high, milk yield ;
3 (humans) choose / cross / breed from, desirable offspring ;
4 continue / repeat, for (several) generations ;
5 allele(s) for, high milk yield / desired trait, passed on ;
6 increase in frequency of, good / best / high yield, allele(s) ;
7 artificial insemination / A.I. ;
8
AVP ;

(b) Explain what is meant by disruptive selection.

Ans: any three from:
1 mean / mode / median / average / intermediate, selected against / not favoured / less fit / die ;
2 (both) extremes, selected for / favoured / more fit / survive / reproduce ;
3 bimodal distribution ;
4 diversifying selection ;
5 causes / maintains, polymorphism / genetic diversity ;

Section B

9 (a) Describe how you would carry out an investigation to calculate the effect of substrate
concentration on the rate of respiration of yeast in anaerobic conditions using a redox
indicator, such as DCPIP or methylene blue.

Ans: any eight from:
1 vary / different / range of, glucose / fructose, concentrations ;
2 control / same, volume of yeast (suspension) ;
3 control / same, volume of glucose (solution) ;
4 control / same, volume / concentration, of indicator (solution) ;
5 control temperature (between 20–45oC) ;
6 oil layer to exclude, air / oxygen ;
7 indicator turns from blue (start / oxidised) to colourless (end / reduced) ;
8 due to, hydrogen atoms / H+ / electrons, from, glycolysis / dehydrogenation / respiration ;
9 time how long it takes for colour change OR fix time interval and measure with colorimeter ;
10 repeat experiment ;
11 calculate mean(s) ;
12 graph with glucose concentration (on
x axis) and, time taken / % absorbance / % transmission / rate (on
y) ;
13 rate is 1/t ;
14
AVP ; 

(b) Compare respiration in anaerobic conditions in mammalian tissue and in yeast cells.

Ans: any seven from:
both
1 occur in cytoplasm ;
2 (only) involve glycolysis ;
3 make, 2 (net) / small amount of, ATP ;
4 regenerate NAD (from NADH) ;
comparison
mammalian tissue yeast cells
5 H acceptor / reduction of pyruvate / pyruvic acid ethanal
6 (end) product lactate / lactic acid ethanol
7 no, CO2 /decarboxylation CO2
8 (enzyme) lactate dehydrogenase ethanol dehydrogenase
9 steps / stages one / 1 two / 2
10 (process) reversible irreversible 

10 (a) Explain how speciation can occur due to geographical separation.

Ans: any eight from:
1 named geographical barrier ; e.g. river / mountain / sea / lake
2 population, separated / divided (into two) ;
3 no, mating / breeding / gene flow, between, populations / groups ;
4 different (named), selection pressures / environments / conditions ;
5 different mutations ;
6 individuals with beneficial alleles, are selected for / survive / reproduce / have a selective advantage ;
7 genetic drift / founder effect ;
8 change in, allele frequency / gene pool ;
9 over a long time / many generations ;
10 can no longer (successfully), interbreed / reproduce / produce fertile offspring, with, old / original / other, population /
species ;
11 allopatric (speciation) ; 

(b) Using examples, explain why species may become extinct

Ans: any seven from:
1 environment changes ;
2 species cannot adapt (fast enough) ;
3 climate / temperature / sea level, change ;
4 extinct species / species at risk ;
5 competition / predation / disease, from, new / introduced / alien / invasive, species ;
6 extinct species / species at risk ;
7 (named) habitat, degradation / loss ;
8 extinct species / species at risk ;
9 hunting by humans / poaching / (over)fishing ;
10 extinct species / species at risk ;
11 AVP ;
12 extinct species / species at risk ;

1 The enzyme β-galactosidase catalyses the breakdown of the compound ONPG (o-nitrophenyl
1-D-galactopyranoside) to the compound ONP (o-nitrophenyl), as shown in Fig. 1.1.
ONPG
(colourless substrate)
β-galactosidase
ONP
(yellow product)
+ galactose
Fig. 1.1
As ONP is produced, the colour of the reaction mixture changes to yellow. The intensity of the
yellow colour produced is proportional to the concentration of ONP.
A colorimeter is used to measure the absorbance of the reaction mixture. Absorbance is a measure
of the light absorbed by a coloured solution. In the reaction shown in Fig. 1.1, the more intense the
yellow colour, the higher the absorbance.
(a) A student was provided with a stock solution of the enzyme β-galactosidase. The student
diluted this by a factor of 20 using a buffer solution of pH 8. The student made a final volume
of 10cm3 of dilute β-galactosidase solution.
Describe how the student prepared the 10cm3 of diluted β-galactosidase solution

Ans: 0.5 cm3 of the stock solution ;
idea of adding 9.5 cm3 of buffer / water / filling to the 10 cm3 mark with buffer / water (in a volumetric flask) ;

The student investigated the effect of substrate concentration on the enzyme-catalysed reaction
shown in Fig. 1.1.
(b) The student was provided with:
• the diluted β-galactosidase solution prepared in step (a), which was kept cold until
needed
• a stock solution of 1.0% ONPG made up in a buffered solution of pH 8.0
• a buffer solution of pH 8.0.
The procedure used by the student is outlined in step 1 to step 3.
1. The diluted β-galactosidase solution was mixed with 1.0% ONPG solution.
2. After 2 minutes a colorimeter was used to measure the absorbance of this mixture.
3. Steps 1 and 2 were repeated using different concentrations of ONPG solution.

(i) Suggest why the student used a colorimeter to measure the absorbance rather than
judging the intensity of the colour by eye.

Ans: idea that (result / it, is) quantitative / AW ; ora 

(ii) Identify the independent variable and the dependent variable in this investigation

Ans : independent variable:
concentration of ONPG / substrate ;
dependent variable:
absorbance (on colorimeter) ;

(c) The student decided to investigate the effect of inhibitors on the reaction shown in Fig. 1.1.
The student planned to add an inhibitor, inhibitor X, to reaction mixtures containing different
concentrations of ONPG solution.

(i) Describe a method the student could use to collect the data needed to test the effect of
inhibitor X in reaction mixtures containing different concentrations of ONPG solution.
The description of your method should be set out in a logical way and be detailed enough
for another person to follow.
You should not repeat the details from (a) describing how to dilute the stock solution of
β-galactosidase.

Ans: 

any eight from:
1 ref. to preparing / using at least five concentrations of ONPG
2 stated range of five dilutions, from 1%v down, with % units
3 method of keeping
β-galactosidase / enzyme, cold (prior to use) ;
4 idea of bringing enzyme and substrate to temperature / equilibrating, before mixing ;
5 method of keeping reactants / reaction mixture, at, constant / controlled / stated / standardised, temperature ;
6 same / constant / stated / standardised / known, volume / concentration / mass of,
X / inhibitor ;
7 same / constant / stated / standardised / known, volume of, ONPG / substrate ;
8 same / constant / stated / standardised / known, volume / concentration, of
β-galactosidase / enzyme ;
9 ref. to measuring absorbance / intensity of colour (on colorimeter) after one set time ;
10 additional detail of use of colorimeter ;
11 repeat at least twice / three replicates, and calculate mean ;
12 idea of repeating, whole experiment / range of concentrations, without
X / inhibitor ;
13 named hazard and risk and precaution ;

The student suggested that inhibitor X was acting as a competitive inhibitor.

(ii) On Fig. 1.3, sketch the curve expected if inhibitor X was acting as a competitive inhibitor.
[2]
Vmax is the maximum initial rate of reaction of the enzyme.
The Michaelis-Menten constant, Km, is the substrate concentration at which the initial rate of
reaction is half its maximum value, Vmax.

Ans: 1 curve, to the right of / below, original curve ;
2 curve / line, must meet the plateau of original curve on the graph ;

(iii) Draw on Fig. 1.3 the positions of Vmax and Km of the enzyme when no inhibitor X is
present.

Ans: 1 correct placement of Vmax ;
2 correct placement of K
m ;

(iv) Use your graph to describe the effect of the addition of inhibitor X on the Km of this
enzyme.

Ans:   (reaction with competitive inhibitor / X) has higher / has increased / increases, K
m value (compared to no inhibitor) ;

(d) Acid reflux is a condition where some of the stomach contents are forced back up into the
oesophagus (gullet). The main symptom is a burning pain in the oesophagus due to the
acidic contents of the stomach. Acid reflux that happens more than twice a week is called
gastroesophageal reflux disease (GERD).
Two main types of drug are used to treat GERD:
• proton pump inhibitors (PPIs)
• H2 receptor antagonists (H2RAs)

Scientists carried out trials to investigate the effect of these drugs on the relief of acid reflux
symptoms in people suffering with GERD.
• 200 people were randomly divided into two groups, A and B.
• People in group A were given the PPI medication.
• People in group B were given the H2RA medication.
• The trial lasted for 16 weeks.
• In weeks 4, 8, 12 and 16, the people were asked to score their symptoms, using the
scale shown in Table 1.1.
Table 1.1
scale symptom
1 none
2 minimal
3 mild
4 moderate
5 moderately severe
6 severe
7 very severe
• Mean values for each of the treatments A and B were calculated for weeks 4, 8, 12
and 16.
• People with a score of 1 were classified as showing removal of symptoms.
• People with a score of 2 or 3 were classed as showing improvement of symptoms.
The results of these trials are shown in Fig. 1.4 and Fig. 1.5.

The scientists analysed the data and concluded that PPIs should be used to treat acid reflux
rather than H2RAs.
With reference to the data in Fig. 1.4 and Fig. 1.5, discuss the conclusion that PPIs should be
used to treat acid reflux.

Ans:  assume ref. to PPI unless specific treatment stated
any three from:
1 idea that PPI overall is more effective / is immediately effective, on improvement / removal (of symptoms) ;
2
either
at 12 and 16 weeks / later, overlap of error bars means difference (in effect between PPI and H2RA) is not significant ;
or
at 4 and 8 weeks / earlier, no overlap of error bars – suggesting difference (in effect, between PPI and H2RA) may be,
significant ;
3 no statistical test to compare groups / t test, done (to test for significance) ;
4 idea that
H2RA is catching up / may be better in the long term ;
5 AVP ; ; any two from:
• less than 50% patients reported no symptoms
• no information on severity of symptoms pre-trial
• no data on side effects
• size of the two test groups not given
• no data on long term effects (where H2RA may be better) 

2 Inheritance of flower colour and flower position in pea plants are controlled by two genes.
• Gene P/p controls flower colour. Allele P for purple flowers is dominant to allele p for
white flowers.
• Gene A/a controls flower position. Allele A for flowers growing from the side of the shoot
(axial position) is dominant to allele a for flowers growing at the end of the shoot (terminal
position), as shown in Fig. 2.1.

A biologist predicted that, if the genes are on different chromosomes, the ratio of the phenotypes
of the F2 generation would be 9:3:3:1.
The biologist carried out a breeding experiment.
• Plants homozygous for white flowers and axial position were crossed with plants
homozygous for purple flowers and axial position.
• All the F1 plants had purple, axial flowers.
• The F1 plants were crossed with each other.
Table 2.1 shows the results for the F2 generation.
Table 2.1
F2 phenotype frequency
purple, axial flowers 1756
purple, terminal flowers 653
white, axial flowers 702
white, terminal flowers 234
total 3345

(a) The chi-squared test (χ2 test) was used to analyse the data in Table 2.1.
(i) State one reason why the chi-squared test (χ2 test) was used.

Ans: idea of compare / AW, observed and expected (data / results / ratios / phenotypes)
or
data is categoric / discrete / discontinuous / nominal ; 

(ii) State the null hypothesis that the biologist would use for this test

Ans:  idea of no (significant) difference between the observed and the expected (results)
or
idea of difference between observed and expected (results) is due to chance ;

(iii) Complete Table 2.2 and calculate the value of χ2 for the results of the F2 generation.
The equation for the calculation of χ2 is:
χ2 = ∑ (O – E)
2
E

Ans:

offspring phenotype O E ( )2 O E
E
−
purple axial flowers 1756 1882 (A 1881)
(1881.56250)
8.44
(8.379175)
purple, terminal flowers 653 627
(627.18750)
1.08
(1.062338)
white, axial flowers 702 627
(627.18750)
8.97
(8.923823)
white, terminal flowers 234 209 ;
(209.06250)
2.99 ;
(2.974608)
2 χ = 21.48 ;
(21.339944)
2 χ = 21.33 / 21.34 / 21.48 for mp3 (to allow for using calculator values or rounded values)

(iv) State the critical value at p < 0.05 for this χ2 test

Ans:  7.82 ;

(v) Use your calculated value of chi-squared (χ2) to:
• explain whether the null hypothesis should be accepted or rejected

• suggest a conclusion the biologist could make about the inheritance of the genes

Ans: 

apply ecf as needed
1 reject (null hypothesis) and state that, the calculated value / 2 χ , is higher than, 7.82 / the critical value / the critical value
they give in (iv) from the table ;
2 If null hypothesis rejected:
(between O and E): there is a significant difference / the difference is not due to chance
idea that there is a less than 5% probability that the difference is due to chance / there is a more than 95% probability
that the difference is not due to chance ;
3 From mp1 If null hypothesis rejected or mp2 If there is a significant difference
/ AW stated
idea that: genes are on the same chromosome / are not on different chromosomes / are linked / show autosomal
linkage / do not assort independently ;
A idea of, ‘may be / can be’, throughout